Answer:
[OH⁻] = 0.000324619M
pH = 10.51
[CN⁻] = 5.175x10⁻³M
Explanation:
Full question says:
Calculate the OH- concentration and pH of a 5.5×10-3M aqueous solution of sodium cyanide, NaCN. Finally, calculate the CN- concentration.
Ka (HCN) = 4.9×10-10
Sodium cyanide, is in equilibrium with water as follows:
NaCN(aq) + H₂O(l) ⇄ HCN(aq) + OH⁻(aq)
Where Kb (Kb = Kw / Ka = 1x10⁻¹⁴ / 4.9x10⁻¹⁰ = 2.04x10⁻⁵) is:
2.04x10⁻⁵ = [HCN] [OH⁻] / [NaCN]
In equilibrium, ann amount of HCN and OH⁻ is produced = X and the molarity of NaCN is: 5.5x10⁻³M - X
Replacing in Kb expression:
2.04x10⁻⁵ = [X] [X] / [5.5x10⁻³M - X]
1.12x10⁻⁷- 2.04x10⁻⁵X = X²
1.12x10⁻⁷- 2.04x10⁻⁵X - X² = 0
Solving for X:
X = -0.0003 → False solution. There is no negative concentrations
X = 0.000324619
As [OH⁻] = X.
pOH = -log [OH⁻]
pOH = 3.489
pH = 14 -pOH
And as molarity of NaCN = [CN⁻] = 5.5x10⁻³M - X
